[LeetCode] 703. Kth Largest Element in a Stream_Easy tag: Heap

Design a class to find the kth largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth distinct element.

Implement KthLargest class:

• KthLargest(int k, int[] nums) Initializes the object with the integer k and the stream of integers nums.
• int add(int val) Returns the element representing the kth largest element in the stream.

 

Example 1:

Input
["KthLargest", "add", "add", "add", "add", "add"]
[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
Output
[null, 4, 5, 5, 8, 8]

Explanation
KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
kthLargest.add(3);   // return 4
kthLargest.add(5);   // return 5
kthLargest.add(10);  // return 5
kthLargest.add(9);   // return 8
kthLargest.add(4);   // return 8

 

Constraints:

• 1 <= k <= 104
• 0 <= nums.length <= 104
• -104 <= nums[i] <= 104
• -104 <= val <= 104
• At most 104 calls will be made to add.
• It is guaranteed that there will be at least k elements in the array when you search for the kth element.

 

Ideas: 利用heap， 先建立一个min heap, 然后每次保证heap的size <= k, 即可。

class KthLargest:

    def __init__(self, k: int, nums: List[int]):
        self.heap = nums
        self.k = k
        heapq.heapify(self.heap)
        
    def add(self, val: int) -> int:
        heapq.heappush(self.heap, val)
        while len(self.heap) > k:
            heapq.heappop(self.heap)
        return self.heap[0]

 

2. 在initial的时候就将其做成size 最多为k的heap， 后面每次add val的时候，如果val 比heap[0] 小就不用管了， 每次add可以节省logk的时间复杂度。

class KthLargest:

    def __init__(self, k: int, nums: List[int]):
        self.heap = nums
        self.k = k
        heapq.heapify(self.heap)
        while len(self.heap) > self.k:
            heapq.heappop(self.heap)

    def add(self, val: int) -> int:
        if len(self.heap) < self.k:
            heapq.heappush(self.heap, val)
        elif val > self.heap[0]:
            heapq.heapreplace(self.heap, val)
        return self.heap[0]